Question

In solid carbon monoxide, each CO molecule has two possible orientations: CO or OC. Assuming that these orientations are completely random (not quite true but close), calculate the residual entropy of a mole of carbon monoxide.

Short answer

The residual entropy for a carbon monoxide is calculated to be$5.76J{K}^{-1}$.

Step-by-step solution

Given

Carbon monoxide is the given solid. CO is its chemical formula. Each molecule can be arranged in two different ways: CO or OC.

Calculation

Multiplicity for $N$molecules is given as:

$\Omega =2^N\dots \dots \dots ..\left(1\right)$

And Residual entropy is given as:

$S_{\text{residual}}=k\ln \Omega \dots \dots \dots \text{(2)}$

Where, $k$is Boltzmann constant.

In the given case, $N=6.022\times {10}^{23}$

By substituting this value in equation (1), we get,

$\Omega =2^{6.022\times {10}^{23}}$

Now, by substituting this value of $\Omega$and $k=1.38\times {10}^{-23}J{K}^{-1}$in equation (2), we get,

$$\begin{gathered} S_{\text{residual}}=\left(1.38\times {10}^{-23}\right)\ln \left(2^{6.022\times {10}^{23}}\right) \\ {S}_{\text{residual}}=\left(1.38\times {10}^{-23}\right)\left(6.022\times {10}^{23}\right)\ln 2 \\ {S}_{\text{residual}}=5.76{\mathrm{JK}}^{-1} \end{gathered}$$

Final answer

Hence, the residual entropy is calculated to be$5.76J{K}^{-1}$.